/*
To the Max
Time Limit: 1 Second      Memory Limit: 32768 KB

Problem

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.

As an example, the maximal sub-rectangle of the array:

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2

is in the lower left corner:

9 2
-4 1
-1 8

and has a sum of 15.

The input consists of an N x N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N 2 integers separated by whitespace (spaces and newlines). These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].


Output

Output the sum of the maximal sub-rectangle.


Example

Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

Output

15 
*/

#include <iostream>
using namespace std;


int N=0;
int **M;

int calc_matrix(int r,int c,int er,int ec) {
	int ret=0;
	for (int i=r;i<=er;i++)
		for (int j=c;j<=ec;j++) 
			ret += M[i][j];
	return ret;
}

int cal(int r,int c) {
	int max = M[r][c],t;
	for (int i=r;i<N;i++)
		for (int j=c;j<N;j++) {
			t=calc_matrix(r,c,i,j);
			if(max<t) max = t;
		}
	return max;
}

int main(){
	cin>>N;
	M = new int* [N];
	for (int i=0;i<N;i++) 
		M[i]=new int[N];

	int t,count=0;
	while (cin>>t) {
		M[count/N][count%N] = t;
		++ count;
	}

	int max = M[0][0];
	for (int i=0;i<N;i++)
		for (int j=0;j<N;j++) 
			if ((t=cal(i,j))>max)
				max = t;

	cout<<max<<endl;
}
